The Physics Language – Physicspupil

September 5, 2024September 5, 2024

202409051606 Pastime Exercise 012

About the figure below, tell some stories as probable as probable can be.

Roughwork.

Ignoring gravitational effect, denote the initial and the final velocity before and after impact by $\mathbf{u}=(u_x,u_y)$ and $\mathbf{v}=(v_x,v_y)$ . Taking the positive an up and a right, then $u_x,u_y,v_x<0$ and $v_y>0$ . Under conservation of momentum,

$\begin{aligned} m|\mathbf{u}|+M|\mathbf{U}| & =m|\mathbf{v}|+M|\mathbf{V}| \\ m|(u_x\,\hat{\mathbf{i}}+u_y\,\hat{\mathbf{j}})| + M|(\mathbf{0})| & = m|(v_x\,\hat{\mathbf{i}}+v_y\,\hat{\mathbf{j}})| + M|(\mathbf{0})| \\ |u_x| & = |v_x| \\ |u_y| & = |v_y| \\ \textrm{( } |u_i| & = |v_i| \textrm{ )}\\ \end{aligned}$

and conservation of energy

$\begin{aligned} \frac{1}{2}mu^2 + \frac{1}{2}MU^2 & = \frac{1}{2}mv^2 + \frac{1}{2}MV^2 \\ m(u_x^2+u_y^2) + M(0)^2 & = m(v_x^2+v_y^2) + M(0)^2 \\ u_x^2+u_y^2 & = v_x^2 + v_y^2 \\ \sum u_i^2 & = \sum v_i^2 \\ \end{aligned}$

Therefore

$\begin{aligned} \frac{|u_x|}{|u_y|} & = \frac{|v_x|}{|v_y|} \\ \tan\theta_i & = \tan\theta_r \\ \theta_i & = \theta_r \\ \end{aligned}$

(to be continued)

August 22, 2024

202408221017 Exercise 5.68

A particle moves along the space curve

$\mathbf{r}=e^{-t}\cos t\,\hat{\mathbf{i}}+e^{-t}\sin t\,\hat{\mathbf{j}}+e^{-t}\,\hat{\mathbf{k}}$ .

Find the magnitude of the (a) velocity and (b) acceleration at any time $t$ .

_{Extracted from Spiegel, M. R. (1971). Schaum’s Outline of Theory and Problems of Advanced Mathematics for Engineers and Scientists.}

Roughwork.

Let displacement

$\mathbf{r}=r_x\,\hat{\mathbf{i}}+r_y\,\hat{\mathbf{j}}+r_z\,\hat{\mathbf{k}}$

where

$\begin{aligned} r_x(t) & = e^{-t}\cos t \\ r_y(t) & = e^{-t}\sin t \\ r_z(t) & = e^{-t} \\ \end{aligned}$

then velocity

$\mathbf{v}=v_x\,\hat{\mathbf{i}}+v_y\,\hat{\mathbf{j}}+v_z\,\hat{\mathbf{k}}$

where

$\begin{aligned} v_x & = \frac{\mathrm{d}r_x}{\mathrm{d}t} \\ & =\frac{\mathrm{d}}{\mathrm{d}t}(e^{-t}\cos t) \\ v_x(t)& = -e^{-t}\sin t-e^{-t}\cos t \\ v_y & = \frac{\mathrm{d}r_y}{\mathrm{d}t} \\ & =\frac{\mathrm{d}}{\mathrm{d}t}(e^{-t}\sin t) \\ v_y(t)& = e^{-t}\cos t-e^{-t}\sin t \\ v_z & = \frac{\mathrm{d}r_z}{\mathrm{d}t} \\ & =\frac{\mathrm{d}}{\mathrm{d}t}(e^{-t}) \\ v_z(t)& = -e^{-t} \\ \end{aligned}$

then acceleration

$\mathbf{a}=a_x\,\hat{\mathbf{i}}+a_y\,\hat{\mathbf{j}}+a_z\,\hat{\mathbf{k}}$

where

$\begin{aligned} a_x & = \frac{\mathrm{d}v_x}{\mathrm{d}t} \\ & =\frac{\mathrm{d}}{\mathrm{d}t}(-e^{-t}\sin t-e^{-t}\cos t) \\ a_x(t)& = 2e^{-t}\sin t\\ a_y & = \frac{\mathrm{d}v_y}{\mathrm{d}t} \\ & =\frac{\mathrm{d}}{\mathrm{d}t}(e^{-t}\cos t-e^{-t}\sin t) \\ a_y(t)& = -2e^{-t}\cos t\\ a_z & = \frac{\mathrm{d}v_z}{\mathrm{d}t} \\ & =\frac{\mathrm{d}}{\mathrm{d}t}(-e^{-t}) \\ a_z(t)& = e^{-t} \\ \end{aligned}$

Cheat.

$\mathbf{r}(t) = \underbrace{e^{-t}}_{\textrm{(1)}}\underbrace{(\cos t,\sin t, 1)}_{\textrm{(2)}}$

where

$\begin{aligned} \textrm{(1):}&\enspace \textrm{radial centripetal} \\ \textrm{(2):}&\enspace \textrm{circumferential anticlockwise} \\ \end{aligned}$

_{By courtesy of WolframAlpha}

In a cylindrical coordinate system, the position of a particle can be written as

$\mathbf{r}=\rho\,\hat{\boldsymbol{\rho}}+z\,\hat{\mathbf{z}}$ .

The velocity of the particle is the time derivative of its position,

$\mathbf{v}=\displaystyle{\frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t}}=\dot{\rho}\,\hat{\boldsymbol{\rho}}+\rho\dot{\varphi}\,\hat{\boldsymbol{\varphi}}+\dot{z}\,\hat{\mathbf{z}}$ ,

where the term $\rho\dot{\varphi}\,\hat{\boldsymbol{\varphi}}$ comes from the Poisson formula

$\displaystyle{\frac{\mathrm{d}\hat{\boldsymbol{\varphi}}}{\mathrm{d}t}}=\dot{\varphi}\,\hat{\mathbf{z}}\times\hat{\boldsymbol{\rho}}$ .

Its acceleration is

$\mathbf{a}=\displaystyle{\frac{\mathrm{d}\mathbf{v}}{\mathrm{d}t}}=(\ddot{\rho}-\rho\dot{\varphi}^2)\,\hat{\boldsymbol{\rho}}+(2\dot{\rho}\dot{\varphi}+\rho\ddot{\varphi})\,\hat{\boldsymbol{\varphi}}+\ddot{z}\,\hat{\mathbf{z}}$ .

_{Kinematics in Wikipedia on Cylindrical coordinate system}

Stop wandering! You are facing up to

$\begin{aligned} v=|\mathbf{v}|=\sqrt{v_x^2+v_y^2+v_z^2} & = \cdots \\ a=|\mathbf{a}|=\sqrt{a_x^2+a_y^2+a_z^2} & = \cdots \\ \end{aligned}$

But then, I’m fond of opting out.

This problem is not to be attempted.

August 21, 2024August 22, 2024

202408211136 Exercise 14.C.10

A stone is thrown vertically upwards so that its height, $s$ metres, above the ground, after $t$ seconds is given by $s=40t-5t^2$ . Find

(a) its velocity after $2\,\mathrm{s}$ ;
(b) its height above the ground when it is momentarily at rest;
(c) its initial velocity;
(d) its velocity when it is $15\,\mathrm{m}$ above the ground, giving your answer correct to the nearest $\mathrm{m\,s^{-1}}$ .

_{Extracted from K. S. Teh & C. Y. Loh. (2007). New Syllabus Additional Mathematics (8e)}

Roughwork.

Initially (i.e., at $t=0$ ), the stone is

$s(t=0)=40(0)-5(0)^2=0\,\mathrm{m}$

above the ground, i.e., at ground level $s=0$ .

$\begin{aligned} & s = 40t-5t^2 = t(40-5t) \\ \Rightarrow & \begin{cases} s =0 \Leftrightarrow t = \{0\}\cup\{ 8\} \\ s > 0 \Leftrightarrow t\in (0,8) \\ s < 0 \Leftrightarrow t<0\enspace\textrm{(rej.)}\enspace\textrm{\scriptsize{OR}}\enspace t>8 \\ \end{cases} \\ \end{aligned}$

Take upward positive. The (vertical) displacement of the stone is

$\begin{aligned} \mathbf{s}(t) & =s(t)\,\hat{\mathbf{j}} \\ & = (40t-5t^2)\,\hat{\mathbf{j}} \\ \end{aligned}$

its (vertical) instantaneous velocity

$\begin{aligned} \mathbf{v}(t) & = v(t)\,\hat{\mathbf{j}} \\ & = \frac{\mathrm{d}s(t)}{\mathrm{d}t}\,\hat{\mathbf{j}} \\ & = (40t-5t^2)'\,\hat{\mathbf{j}} \\ & = (40-10t)\,\hat{\mathbf{j}} \\ \end{aligned}$

and its (vertical) instantaneous acceleration

$\begin{aligned} \mathbf{a}(t) & = a(t)\,\hat{\mathbf{j}} \\ & = \frac{\mathrm{d}v(t)}{\mathrm{d}t}\,\hat{\mathbf{j}} \\ & = (40-10t)'\,\hat{\mathbf{j}} \\ \mathbf{a} & = -10\,\hat{\mathbf{j}} \\ \end{aligned}$

so are the answers to

(a) $\mathbf{v}(2)$
(b) $t'\in\{ t:v(t)=0\}\enspace\rightsquigarrow\enspace s(t')$
(c) $\mathbf{v}(0)$
(d) $t'\in\{ t:\mathbf{s}(t)=+15\,\hat{\mathbf{j}}\} \enspace\rightsquigarrow\enspace \mathbf{v}(t')$

This problem is not to be attempted.

August 20, 2024August 20, 2024

202408201533 Example 13.8

Let $\displaystyle{x=\frac{y-2}{y+2}}$ .

(a) Find $\displaystyle{\frac{\mathrm{d}y}{\mathrm{d}x}}$ in terms of $y$ .
(b) Make $y$ the subject of the given function. Hence, find $\displaystyle{\frac{\mathrm{d}y}{\mathrm{d}x}}$ in terms of $x$ .
(c) Are the answers obtained in (a) and (b) different? Explain your assertion.

_{Extracted from S. W. Li et al. (2002). New Progress in Additional Mathematics Book 3}

Roughwork.

$\begin{aligned} x & = \frac{y-2}{y+2} \\ x & = \frac{y+2-4}{y+2} \\ \textrm{Eq. (1):}\quad x=f(y) & = 1-\frac{4}{y+2} \\ \frac{4}{y+2} & = 1-x \\ \frac{4}{1-x} & = y+2 \\ \textrm{Eq. (2):}\quad y=g(x) & = \frac{4}{1-x} - 2 \\ \end{aligned}$

domain: what can go into a function
codomain: what may possibly come out of a function
range: what actually comes out of a function

_{Kevin Sookocheff on Range, Domain, and Codomain (2018)}

Write, for function $f$ ,

$\begin{aligned} \textrm{domain: }& \mathbb{R}\backslash\{ -2\} \\ \textrm{codomain: }& \mathbb{R} \\ \textrm{range: }& \mathbb{R} \\ \end{aligned}$

and for function $g$ ,

$\begin{aligned} \textrm{domain: }& \mathbb{R}\backslash\{ 1\} \\ \textrm{codomain: }& \mathbb{R} \\ \textrm{range: }& \mathbb{R} \\ \end{aligned}$

We see $f$ discontinuous at $-2$ ; and $g$ , at $1$ ; whereas non-differentiable.

(to be continued)

August 19, 2024August 20, 2024

202408191743 Geometry 0

$\forall$ : for all/any/each/every
$\exists$ : there exist(s)/is/are

If, in any event $(x,y,z,t)$ :

i. no two points $a$ and $b$ occupy the same position. I.e.,

$\forall\, t\, \nexists\, a(\neq b)\enspace\textrm{s.t. }(x_a,y_a,z_a)=(x_b,y_b,z_b)$

ii. none any point $a$ occupies more than one position. I.e.,

$\forall\,a,t\,\exists\, !\mathbf{r}\enspace\textrm{s.t. }a\stackrel{\mathbf{r}(t)}{\longmapsto}(x_a,y_a,z_a)$

such points are said to be mass particles; or else massless particles.

(to be continued)

May 10, 2024June 13, 2024

202405101842 Exercise 16.67

The cross product of a vector with itself is

A. the vector $\mathbf{i}^2+\mathbf{j}^2+\mathbf{k}^2$ .
B. the vector $\mathbf{i}+\mathbf{j}+\mathbf{k}$ .
C. the zero vector.
D. the scalar quantity $1$ .
E. the scalar quantity $0$ .

_{Extracted from Stan Gibilisco. (2006). Technical Math Demystified.}

Erratum.

Option A should be a scalar.

Roughwork.

Most commonly, it is the three-dimensional Euclidean space $\mathbf{E}^3$ (or $\mathbb{E}^3$ ) that models physical space $\mathbf{R}^3$ (or $\mathbb{R}^3$ ).

_{Wikipedia on Three-dimensional space}

The Euclidean metric is given by

$\mathbf{G}=(g_{\alpha\beta})\equiv \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\\end{bmatrix}$

and so line element $\mathrm{d}s$

$\begin{aligned} \mathrm{d}s^2 & = g_{\alpha\beta}\,\mathrm{d}x^\alpha\,\mathrm{d}x^\beta \\ & = (\mathrm{d}x^1)^2 + (\mathrm{d}x^2)^2 + (\mathrm{d}x^3)^2 \\ & = \mathrm{d}x^2+\mathrm{d}y^2+\mathrm{d}z^2 \\ \end{aligned}$

or distance

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$

is equivalent to applying twice the Pythagorean equation

$a^2+b^2=c^2$

to two points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ with respect to the standard basis $(\mathbf{e}_x,\mathbf{e}_y,\mathbf{e}_z)$ :

$\begin{cases} \hat{\mathbf{i}}=\mathbf{e}_x=(1,0,0) \\ \hat{\mathbf{j}}=\mathbf{e}_y=(0,1,0) \\ \hat{\mathbf{k}}=\mathbf{e}_z=(0,0,1) \\ \end{cases}$

of a Cartesian coordinate system. To carry a point from one position to another, a vector $\mathbf{v}$ has both magnitude $v=|\mathbf{v}|$ and direction $\hat{\mathbf{v}}=\frac{\mathbf{v}}{|\mathbf{v}|}$ . Scalar multiplication $c\mathbf{v}$ will scale its magnitude by some factor $c$ (the scalar) without change in its direction. I.e.,

$\begin{aligned} \mathbf{u} & \stackrel{\mathrm{def}}{=}c\mathbf{v} \\ u & = |\mathbf{u}| = |c\mathbf{v}| = c|\mathbf{v}|\\ & = cv \\ \hat{\mathbf{u}} & = \frac{\mathbf{u}}{|\mathbf{u}|} = \frac{c\mathbf{v}}{c|\mathbf{v}|} = \frac{\mathbf{v}}{|\mathbf{v}|} \\ & = \hat{\mathbf{v}} \\ \end{aligned}$

Vector multiplication is for whose vectors be the products scalar or vector. For instance, we have

Dot product (aka scalar product) of two vectors $\mathbf{u}=(u_x,u_y,u_z)$ and $\mathbf{v}=(v_x,v_y,v_z)$

$\begin{aligned} \mathbf{u}\cdot\mathbf{v} & = \sum_{x,y,z}u_iv_i \\ & = u_xv_x+u_yv_y+u_zv_z \\ & = |\mathbf{u}| |\mathbf{v}|\cos\theta \\ \end{aligned}$

as a scalar quantity; but their cross product (aka vector product)

$\begin{aligned} \mathbf{u}\times\mathbf{v} & = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ u_x & u_y & u_z \\ v_x & v_y & v_z \\ \end{vmatrix} \\ & = (u_yv_z-u_zv_y, u_zv_x-u_xv_z, u_xv_y-u_yv_x) \\ & = |\mathbf{u}||\mathbf{v}|\sin\theta\,\hat{\mathbf{n}} \\ \end{aligned}$

a vector quantity. Added on are tensor product $\mathbf{u}\otimes\mathbf{v}$ , wedge product $\mathbf{u}\wedge\mathbf{v}$ , and more.

This problem is not to be attempted.

May 9, 2024May 10, 2024

202405091824 Pastime Exercise 011

About the graph below, tell some stories as probable as probable can be.

Roughwork.

Assuming linear (/rectilinear) motion in a single direction.

Assuming uniform acceleration, there are two cases: i. zero acceleration and ii. non-zero (constant) acceleration; in the former velocity being (a) constant and the latter (b) non-constant.

Assuming a flat spacetime metric of which the spatial part does not expands with the temporal part.

Imagine a man walking along a straight line from point $(x_1,y_1)$ to point $(x_2,y_2)$ . From

$\displaystyle{\textrm{Speed }(v)=\frac{\textrm{Distance }(s)}{\textrm{Time }(t)}}$

as the path is fixed, i.e., $\Delta s=\textrm{Const.}$ , we see speed $v$ and time $t$

$\begin{aligned} v\uparrow\enspace \Leftrightarrow \enspace& t\downarrow \\ v\downarrow\enspace\Leftrightarrow\enspace& t\uparrow \\ \end{aligned}$

in an inverse relationship. As the man keeps his own fair pace and makes himself a good timekeeper, we can treat speed $v$ as an independent variable, and time $t$ a dependent variable.

$\textrm{\scriptsize{CASE} \textbf{\texttt{(a)}}}$ : velocity being constant

Parameterize the Cartesian equation $y=mx+c$ by the parameter $t'$ (*as distinguished from the natural/unit-speed/arc-length parameter time $t$ ) so as to write a set of parametric equations:

$\begin{aligned} & \begin{cases} s_x(t')=t' \\ s_y(t')=mt'+c \\ \end{cases} \\ & \\ & \begin{cases} v_x(t') =s_x'(t') = 1 \\ v_y(t') = s_y'(t') = m \\ \end{cases} \\ \end{aligned}$

$\begin{aligned} v^2(t') & = v_x^2(t')+v_y^2(t') \\ v(t') & = \displaystyle{\sqrt{\bigg(\frac{\mathrm{d}s_x(t')}{\mathrm{d}t'}\bigg)^2 +\bigg(\frac{\mathrm{d}s_y(t')}{\mathrm{d}t'}\bigg)^2}} \\ |\mathbf{v}(t')| & = \sqrt{1+m^2} \\ t & = \int_{0}^{t'}|\mathbf{v}(t)|\,\mathrm{d}t \\ & = (\sqrt{1+m^2})t'\\ \end{aligned}$

$\textrm{\scriptsize{CASE} \textbf{\texttt{(b)}}}$ : velocity being non-constant

The man begins with initial speed $v_1$ at start point $(x_1,y_1)$ and ends with final speed $v_2$ at finish point $(x_2,y_2)$ . We have

$\displaystyle{\textrm{Acceleration }(a)=\frac{\textrm{Change in speed }(v-u)}{\textrm{Time }(t)}}$

Write by SUVAT equations of motion:

$\begin{aligned} \mathbf{u} & = (u_x,u_y) \\ \mathbf{v} & = (v_x,v_y) \\ \mathbf{a} & = (a_x,a_y) \\ & = \bigg( \frac{v_x-u_x}{t}, \frac{v_y-u_y}{t}\bigg) \\ \mathbf{s} & = (s_x,s_y) \\ & = \bigg( u_xt+\frac{1}{2}a_xt^2, u_yt+\frac{1}{2}a_yt^2\bigg) \\ & = \bigg(\frac{1}{2}(u_x+v_x)t, \frac{1}{2}(u_y+v_y)t\bigg) \\ & = (x_2-x_1,y_2-y_1) \\ \end{aligned}$

Is this time $t$ also a natural parameter?

For a given parametric curve, the natural parametrization is unique up to a shift of parameter.

_{Wikipedia on Differentiable curve}

(to be continued)

May 8, 2024May 8, 2024

202405081402 Pastime Exercise 010

About the graph below, tell some stories as probable as probable can be.

Roughwork.

We naturally assume there are no forms of negative energy. That is, kinetic energy $\textrm{KE: } K\geqslant 0$ , potential energy $\textrm{PE: }U\geqslant 0$ , and (total) mechanical energy $K+U=E=\textrm{Const.}\geqslant 0$ .

The graph is divided into Zone ①, Zone ②, and Zone ③.

$\begin{aligned} V(x) & = \begin{cases} +\infty & \textrm{for }x\leqslant 1 \\ 2 & \textrm{for }1<x\leqslant 2 \\ -2|x-3|+4& \textrm{for }2\leqslant x\leqslant 5 \\ \frac{1}{2}(x^2-10x+25) & \textrm{for }5\leqslant x\\ \end{cases} \\ \end{aligned}$

For any conservative system, total energy $E$ takes the form of a horizontal line $y=\textrm{Const.}$ as in a graph. Take $E=2$ as an example:

$\begin{aligned} & \qquad E: y=2 \\ & \Longrightarrow \begin{cases} (E-U=)K> 0\Rightarrow x\in (4,7) \\ (E-U=)K = 0\Rightarrow x \in (1,2]\cup\{ 4\}\cup\{ 7\} \\ (E-U=)K< 0\Rightarrow x\in (-\infty ,1]\cup (2,4)\cup (7,\infty) \\ \end{cases} \\ \end{aligned}$

$\begin{aligned} & \qquad K=\frac{1}{2}mv^2 = \frac{p^2}{2m} \\ & \Longrightarrow \begin{cases} K(x)>0\Rightarrow \textrm{a particle moves at }x \\ K(x)=0\Rightarrow\textrm{a particle stays at }x \\ K(x)<0\Rightarrow \textrm{no particle exists at }x \\ \end{cases} \\ \end{aligned}$

(to be continued)

May 7, 2024May 7, 2024

202405071349 Exercise 13.2.1

A right pyramid, $12\,\mathrm{cm}$ high, stands on a rectangular base $6\,\mathrm{cm}$ by $10\,\mathrm{cm}$ . Calculate (a) the length of an edge of the pyramid; (b) the angles the triangular faces made with the base; (c) the volume of the pyramid.

_{Extracted from A. Godman & J. F. Talbert. (1973). Additional Mathematics Pure and Applied in SI Units.}

Roughwork.

Commit my visualization to drawing.

A right pyramid is a pyramid where the base is circumscribed about the circle and the altitude of the pyramid meets at the circle’s center.

_{Wikipedia on Pyramid (geometry)}

The pyramid above has a polygonal base, here the rectangle $QRST$ , and an apex $P$ , here the common vertex of triangles $\triangle PTQ$ , $\triangle PQR$ , $\triangle PRS$ , and $\triangle PST$ . The altitude is based on the origin $O$ . To suit our coordinates to this problem, we write

$\begin{aligned} P & =P(0,0,12) \\ Q & =Q(3,5,0) \\ R & =R(-3,5,0) \\ S & =S(-3,-5,0) \\ T & =T(3,-5,0) \\ \end{aligned}$

such that

$\begin{aligned} a & = c = 10 \\ b & = d = 6 \\ e & = f = g = h \\ & = \surd \big\{ (12)^2 + \big( \sqrt{(6)^2+(10)^2}/2\big)^2 \big\} \\ & = \sqrt{178} \\ & = 13.3417\quad\textrm{(4 d.p.)} \\ \end{aligned}$

For the edges of its base, write

$\begin{aligned} a=\overline{TQ} & :\begin{cases} x=3 \\ |y|\leqslant 5 \\ z=0 \\ \end{cases} \\ b=\overline{QR} & :\begin{cases} |x|\leqslant 3 \\ y=5 \\ z=0 \\ \end{cases} \\ c=\overline{RS} & :\begin{cases} x=-3 \\ |y|\leqslant 5 \\ z=0 \\ \end{cases} \\ d=\overline{ST}& :\begin{cases} |x|\leqslant 3 \\ y=-5 \\ z=0 \\ \end{cases} \\ \end{aligned}$

and for, the lateral, edge $e=\overline{PQ}$ :

$\begin{aligned} \frac{x-3}{0-3} & = \frac{y-5}{0-5} = \frac{z-0}{12-0} \\ \end{aligned}$

edge $f=\overline{PR}$ :

$\begin{aligned} \frac{x-(-3)}{0-(-3)} & = \frac{y-5}{0-5} = \frac{z-0}{12-0} \\ \end{aligned}$

edge $g=\overline{PS}$ :

$\begin{aligned} \frac{x-(-3)}{0-(-3)} & = \frac{y-(-5)}{0-(-5)} = \frac{z-0}{12-0} \\ \end{aligned}$

and edge $h=\overline{PT}$ :

$\begin{aligned} \frac{x-3}{0-3} & = \frac{y-(-5)}{0-(-5)} = \frac{z-0}{12-0} \\ \end{aligned}$

For lateral surface $A$ enclosed by edges $a$ , $h$ , and $e$ , write

$A(x,y,z):\begin{cases} (x,y,z)\in [0,3]\times [-5,5]\times [0,12] \\ \textrm{s.t. }\displaystyle{\frac{|y|}{5}\leqslant \frac{x}{3}=1-\frac{z}{12}} \\ \end{cases}$

for lateral surface $B$ by edges $b$ , $e$ , and $f$ , write:

$B(x,y,z):\begin{cases} (x,y,z)\in [-3,3]\times [0,5]\times [0,12] \\ \textrm{s.t. }\displaystyle{\frac{|x|}{3}\leqslant \frac{y}{5}=1-\frac{z}{12}} \\ \end{cases}$

for lateral surface $C$ by edges $c$ , $f$ , and $g$ , write:

$C(x,y,z):\begin{cases} (x,y,z)\in [-3,0]\times [-5,5]\times [0,12] \\ \textrm{s.t. }\displaystyle{\frac{|y|}{5}\leqslant -\frac{x}{3}=1-\frac{z}{12}} \\ \end{cases}$

for lateral surface $D$ by edges $d$ , $g$ , and $h$ , write:

$D(x,y,z):\begin{cases} (x,y,z)\in [-3,3]\times [-5,0]\times [0,12] \\ \textrm{s.t. }\displaystyle{\frac{|x|}{3}\leqslant -\frac{y}{5}=1-\frac{z}{12}} \\ \end{cases}$

and for base $E$ by edges $a$ , $b$ , $c$ , and $d$ :

write

$E(x,y,z):\begin{cases} |x|\leqslant 3 \\ |y|\leqslant 5 \\ z=0 \\ \end{cases}$

This problem is not to be attempted.

March 1, 2024March 1, 2024

202403011713 Revision Paper IV Q8

Ball $A$ (mass $3\,\mathrm{kg}$ ) is travelling due east in a straight line with speed $5\,\mathrm{m\,s^{-1}}$ when it collides with ball $B$ (mass $4\,\mathrm{kg}$ ) which was at rest. After the collision both balls move east with relative speed $2.5\,\mathrm{m\,s^{-1}}$ . Find their separate speeds.

_{Extracted from A. Godman & J. F. Talbert. (1973). Additional Mathematics Pure and Applied in SI Units.}

Roughwork.

Take eastward positive.

$\begin{aligned} \textrm{CoM:}\qquad p_{\textrm{initial}} & = p_{\textrm{final}} \\ \sum_{i=A}^{B}m_iu_i & = \sum_{i=A}^{B}m_iv_i \\ m_Au_A+m_Bu_B & = m_Av_A+m_Bv_B \\ (3)(5)+(4)(0) & = (3)(v_A)+(4)(v_B) \\ \textrm{Eq. (1):}\qquad\quad 15 & = 3v_A+4v_B \\ \textrm{Eq. (1)':}\qquad\quad v_A & = 5-\frac{4}{3}v_B \\ \end{aligned}$

$\begin{aligned} \textrm{CoE:}\qquad\textrm{KE}_{\textrm{initial}} & = \textrm{KE}_{\textrm{final}} \\ \sum_{i=A}^{B}\frac{1}{2}m_iu_i^2 & = \sum_{i=A}^{B}\frac{1}{2}m_iv_i^2 \\ m_Au_A^2+m_Bu_B^2 & = m_Av_A^2+m_Bv_B^2 \\ (3)(5)^2 + (4)(0)^2 & = (3)(v_A)^2 + (4)(v_B)^2 \\ \textrm{Eq. (2):}\qquad\qquad 75 & = 3v_A^2 + 4v_B^2 \\ \end{aligned}$

$\begin{aligned} 75 & = 3\bigg( 5-\frac{4}{3}v_B\bigg)^2 + 4v_B^2 \\ 25 & = \bigg( 25 - \frac{40}{3}v_B + \frac{16}{9}v_B^2\bigg) + \frac{4}{3}v_B^2 \\ 0 & = v_B\bigg( -\frac{40}{3} + \frac{28}{9}v_B \bigg) \\ v_B & = 0\textrm{ \scriptsize{OR} }\frac{30}{7} \\ v_A & = \bigg[ 5-\frac{4}{3}(0)\bigg] \textrm{ \scriptsize{OR} }\bigg[ 5-\frac{4}{3}\bigg(\frac{30}{7}\bigg)\bigg] \\ & = 5\textrm{ \scriptsize{OR} }-\frac{5}{7} \\ \end{aligned}$

$(v_A,v_B) = \bigg( \displaystyle{-\frac{5}{7} , \frac{30}{7}}\bigg) \qquad \text{}^*\textrm{(5,0) is rejected.}$

This contradicts with $v_B-2.5=v_A>0$ . As the law of conservation of momentum is always observed, herein has energy $\textrm{\scriptsize{NOT}}$ been conserved.

This problem is not to be attempted.